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I have fill a ballon with water. The balloon becomes big for its elasticity. So a potential energy is been stored on the surface of the baloon that is giving pressure on the water inside. Now I need to find that pressure. There is a way to find it using bulk modulus but the problem is I don't know the bulk modulus for the rubber of that ballon.

So I am finding a way to find the pressure.

Also I'm confused if I mesure the weight of the water inside and consider that as force and measure the current area of the ballon and divide then what will be the value? Is that will be the pressure of the ballon! I'm really confused.

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  • $\begingroup$ The balloon is just like a piston in thermodynamic processes, with constant exterior pressure. At the end the balloon is in equilibrium (volume-wise) with its surrounding (air) thus the pressure inside is 1 atmosphere. If you dive with your balloon or take it to high enough altitude (different exterior pressure) you will see the volume change, as the pressure inside will equilibrate with the exterior pressure. $\endgroup$ – Alexander Jun 21 at 7:43
  • $\begingroup$ So you saying that the pressure inside and outside will always be same and when I fill the ballon with water and it's volume increase still the pressure is same as outside. That means I take 100ml water and put in on a glass and take another 100ml water and put it inside a ballon(ballon's volume increase). So both of the water having same pressure? The stress of the baloon causing no extra pressure on the water inside? But the baloon isn't a sphere that all the pressure will cancel out each other and resultant force is zero. But it's a ellipse. I am still confused. $\endgroup$ – Al Faisal Redoy Jun 21 at 8:20
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    $\begingroup$ Your experiment is a clever way to measure the pressure inside the balloon (which is obviously higher than outside), provided the shape of the balloon doesn't change much. To properly calculate the pressure difference between the inside and outside of a spherical balloon (from first principles), you need to measure the biaxial stretching properties of the rubber (force per unit length within the rubber as a function of stretch ratio), and then include this in a differential force balance on a window shaped element in the balloon surface. But be aware that rubber does not obey Hooke's law. $\endgroup$ – Chet Miller Jun 21 at 11:49
  • $\begingroup$ See post #2 in the following thread: physicsforums.com/threads/… $\endgroup$ – Chet Miller Jun 21 at 12:11
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Arrange an apparatus that looks like the following. enter image description here Attach the inflated balloon like this (it's tricky), put some water and release the stopcock.

If the water in the column is too much, the balloon will burst. If the water in the column is too low, the water inside will leak. After some trial and error, you might achieve equilibrium. When there is equilibrium, the pressure inside the balloon will be equal to that due to the weight of the water column and atmospheric pressure. The pressure due to the water column can be obtained by measuring its height. $P = P_{atm} + hdg$

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